Problem: A billiard ball is struck by a cue. It travels $100\,\text{cm}$ before ricocheting off a rail and traveling another $120\, \text{cm}$ into a corner pocket. The angle between the path as the ball approaches the rail and the path after it strikes the rail is $45^{\circ}$. How far is the corner pocket from where the cue initially struck the ball? Do not round during your calculations. Round your final answer to the nearest centimeter.
Answer: Converting the problem into geometrical terms Our problem can be modeled by the following triangle $\triangle ABC$, where we want to find $BC=d$. $45^\circ$ $A$ $B$ $C$ $120\text{ cm}$ $100\text{ cm}$ $d$ Since we are given two side lengths and the angle measure between them, we can use the law of cosines. Using the law of cosines $\begin{aligned} (BC)^2&=(AB)^2+(AC)^2-2AB\!\cdot\! AC\!\cdot\!\cos(A)\\\\ d^2&=120^2+100^2-2\cdot 120\cdot 100\cdot\cos(45^\circ) \gray{\text{Substitute}}\\\\ d&=\sqrt{120^2+100^2-2\cdot 120\cdot 100\cdot\cos(45^\circ)}\\\\ d&\approx 86 \end{aligned}$ The answer The distance from the corner pocket to where the cue initially struck the ball is $86$ centimeters.